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Non-Linear Simultaneous Equations

Date: 12/07/2003 at 15:41:14
From: Alroy
Subject: solving a complex system of simultaneous equations

Solve the simulataneous equations

  ab + c + d = 3
  bc + d + a = 5
  cd + a + b = 2
  da + b + c = 6

where a, b, c, d are real numbers.

I keep arriving at the point where everything cancels out and I am 
left with 0 = 0...


Date: 12/07/2003 at 19:10:02
From: Doctor Schwa
Subject: Re: solving a complex simultaneous equations

Hi Alroy, 

This is a tough one!

To me, it looks useful to combine the first and third equations, 
because they seem to have similar variables.  I'll add them:

  ab + a + b + cd + c + d = 5.

Now one of my favorite tricks comes into play.  Note that

  ab + a + b 

is not a nicely factorable expression.  But 

  ab + a + b + 1 

is -- it's just (a+1)(b+1).  So, add 2 to both sides:

  ab + a + b + 1 + cd + c + d + 1 = 7
  
          (a+1)(b+1) + (c+1)(d+1) = 7

Similarly

          (b+1)(c+1) + (a+1)(d+1) = 13

Adding these together gives

  (a+1)(b+1) + (a+1)(d+1) + (c+1)(b+1) + (c+1)(d+1) = 20

so

  ((a+1) + (c+1))*((b+1) + (d+1)) = 20

Now, if a, b, c, d are yucky numbers, this will still be a big pain.
But, looking at the equations, it seems like you can guess and check
some small integers at this point, integers that multiply to make 20...

In fact, that might be a good approach to take trom the beginning,
simply guessing and checking.

Another possible idea of how to combine the questions is to take 

  (Eq 1) + (Eq 3) - (Eq 2) - (Eq 4)

so that the a+b+c+d subtracts out.  Then you get 

  ab + cd - bc - da = -6

        (a-c)*(b-d) = -6

and again the guessing and checking is pretty easy if the numbers are
small integers.

Well, that's one style, and maybe it works well for you.

Here's another:

  ab + c + d = 3
  bc + d + a = 5
  cd + a + b = 2
  da + b + c = 6

Solve for a, either

  a = 5 - d - bc 

or
  
  a = 2 - b - cd

take your pick.  Substitute one of those into the first equation:

  (5-d-bc)*b + c + d = 3

Now solve something for b, but eliminate a first.  For example,

     d*(5-d-bc) + b + c = 6

  5d - d^2 -bdc + b + c = 6

                      b = (6 - 5d + d^2 - c)/(1 - cd)

You can see that this will eventually solve for all the variables, but 
it will give you perhaps a quite ugly equation for d in the end.  When 
you substitute this b into 

  (5-d-bc)*b + c + d = 3

you'll get quite a messy-looking thing.

So, try something else!

Continuing from the first line of this attempt, where I started 
solving for a:

  5 - d - bc = 2 - b - cd

or 

  3 = (d-b) - c*(d-b)

  3 = (1-c)*(d-b)

But we could also get (d-b) by combining the first and fourth 
equations, eliminating c:

  3 - ab - d = 6 - da - b

or

  -3 = (1-a)*(d-b)

Ah, that's the best approach so far, isn't it?  We already know 1-a
and 1-b are opposites at this point: that's the most concrete 
conclusion I've created in all these attempts.

So 
 
  (1-a) = -(1-b)

    1-a = -1 + b

    a+b = 2

Among other things, that means right away that cd = 0, so c = 0 or 
d = 0, and checking those two cases should help you narrow things down
fairly quickly.

I hope one of these methods helps!  Feel free to write back if you'd
like some more hints.

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Basic Algebra
High School Polynomials

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